`~rai~`

\(A=1+2\sin\left(x+\dfrac{\pi}{4}\right)\\\text{Ta có:}-1\le\sin\left(x+\dfrac{\pi}{4}\right)\le 1\\\Leftrightarrow -2\le2\sin\left(x+\dfrac{\pi}{4}\right)\le 2\\\Leftrightarrow -1\le 1+2\sin\left(x+\dfrac{\pi}{4}\right)\le 3\\\Leftrightarrow -1\le A\le 3.\\+)Min_A=-1\Leftrightarrow \sin\left(x+\dfrac{\pi}{4}\right)=1\\\Leftrightarrow x+\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=-\dfrac{3\pi}{4}+k2\pi.(k\in\mathbb{Z})\\+)Max_A=3\Leftrightarrow \sin\left(x+\dfrac{\pi}{4}\right)=1\\\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=\dfrac{\pi}{4}+k2\pi.(k\in\mathbb{Z})\\\text{Vậy Min}_A=-1\text{ khi x=}-\dfrac{3\pi}{4}+k2\pi;\\\text{Max}_A=3\text{ khi x=}\dfrac{\pi}{4}+k2\pi.(k\in\mathbb{Z})\\B=5+4\sin x\cos x\\\quad=5+2.2\sin x\cos x\\\quad=5+2\sin2x.\\\text{Ta có:}-1\le\sin2x\le 1\\\Leftrightarrow -2\le2\sin2x\le1\\\Leftrightarrow 3\le5+2\sin2x\le7\\\Leftrightarrow 3\le B\le 7.\\+)Min_B=3\Leftrightarrow \sin2x=-1\\\Leftrightarrow 2x=-\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi.(k\in\mathbb{Z})\\+)Max_B=7\Leftrightarrow \sin2x=1\\\Leftrightarrow 2x=\dfrac{\pi}{2}+k2\pi\\\Leftrightarrow x=\dfrac{\pi}{4}+k\pi.(k\in\mathbb{Z})\\\text{Vậy Min}_B=3\text{ khi }x=-\dfrac{\pi}{4}+k\pi;\\\text{Max}_B=7\text{ khi }x=\dfrac{\pi}{4}+k\pi.(k\in\mathbb{Z})\)